The question aroused my interest enough to set up an experiment. I changed the question's parameters in one key aspect: Instead of an LED strip with multiple LEDs in series, I hooked up 3 blue LEDs (Vf = ~2.8 Volts each) in parallel, with a single 100 Ohm resistor to limit current to all 3, to a 0.047 Farad, 5.5 Volt coin type "motherboard supercap".
I know, sharing a resistor is really bad practice, so just use separate resistors for your own experiment.
The supercap was charged from a pair of AA alkaline cells (~3.12 Volts across capacitor after 3 minutes), then the wires to the battery were pulled out.
simulate this circuit – Schematic created using CircuitLab
While the dimming effect was an expected outcome, the results were startling: The LEDs stayed lit at diminishing intensity for over a minute after disconnecting the battery. Here is the video I took of the experiment.
The reason the LEDs stayed lit so much longer than expected is that a typical LED continues to be illuminated down to well under 5% of its nominal current - In the case of the LEDs I used, at around the 1 minute mark they were quite visible, if dim, with a mere 1 mA split between all three.
The LEDs finally dimmed to nothingness after perhaps 15 minutes.
Conclusions:
- A much smaller capacitance than the 0.047 Farad supercapacitor used here would be preferred for the purpose envisaged.
- If one must use a 12 Volt 20 mA LED strip, instead of LEDs in parallel, then a set of 3 of these coin supercaps in series would work: The resultant capacitance of around 0.0157 Farad will provide a dimming duration closer to the OP's target of 2 to 10 seconds, instead of the unbearably long 1 minute dimming observed in the video.
- The reason some previously posted capacitance calculations including my own 0.5 Farad comment were far off the mark is because the reducing current flow due to discharge, i.e. the very dimming effect being sought, was unaccounted for.
- For any comments that might arise about the "unacceptably high" ESR of these motherboard supercaps, it is clear that theory needs to be backed up by practical experimentation, as done for this answer.
The supercapacitor I used is sold for under $2 a pair, including international shipping, on eBay:
Not quite the tens or hundreds of dollars that I, and others, had previously mentioned.
Addeddum thanks to discussion with @DavidKessener:
- If using multiple supercaps in series and charged to a higher voltage for the string, than the individual capacitor's rated voltage, biasing resistors are required to prolong the life of the capacitors. Without these, the capacitors will charge unevenly, and will eventually die faster.
- Based on this Maxwell appnote, and taking a leakage current per capacitor of 10 uA (the actual leakage current of these particular caps is much lower, so even safer), we get a 55 kOhm value for biasing resistors to pass
10 x 10 = 100 uA
, so add 3 new 56k resistors as below, for using a 12 Volt supply and a 12 Volt LED strip
simulate this circuit
Best Answer
looking at the datasheet, your dc power supply is not sufficient.
I would size it about twice the non repetitive peak Amperage for power headroom, but twice its repetitive peak output would be my recommended minimum.
If you go by typical practices, the maximum current draw should be 80% of the power supply so the rectifier filters in the power supply can recover.
But that is only the 1/3 of the issue, the main issue it has is its design. Unbalanced = 100% of power supply noise injected onto the signal.
Now if you create isolated power, or create a ground reference that has the inverse of the power supply noise on the ground, the noise would get subtracted by the output circuit.